# Matrices went in 2x2

• August 21st 2010, 06:52 AM
halbard
Matrices went in 2x2
Another challenge problem, inspired by an earlier question.

Given arbitrary 2x2 matrices $A$ and $B$, prove that

$AB+BA=\beta A+\alpha B+(\gamma-\alpha\beta)I$

where $\alpha$ is the trace of $A$, $\beta$ is the trace of $B$ and $\gamma$ is the trace of $AB$ or of $BA$. $I$ is of course the 2x2 identity matrix.

Although a brute force approach would work, a more elegant solution would be welcome.

Enjoy.

Moderator approved. CB
• August 22nd 2010, 12:35 AM
red_dog
We use the Cayley-Hamilton identity for $A, \ B, \ A+B$

$A^2-\alpha A+\det A\cdot I_2=O_2$

$B^2-\beta B+\det B\cdot I_2=O_2$

$(A+B)^2-Trace(A+B)+\det (A+B)\cdot I_2=O_2$

We have $Trace(A+B)=Trace(A)+Trace(B)=\alpha+\beta$

and $\det(A+B)=\det A+\det B+\gamma-\alpha\beta$

Then

$A^2+B^2+AB+BA-\alpha A-\alpha B-\beta A-\beta B+\det A\cdot I_2+\det B\cdot I_2+(\gamma-\alpha\beta)I_2=O_2$

or $AB+BA=\beta A+\alpha B+(\gamma-\alpha\beta)I_2$
• August 22nd 2010, 01:39 AM
halbard
Quote:

Originally Posted by red_dog
$(A+B)^2-Trace(A+B)+\det (A+B)\cdot I_2=O_2$

Should be

$(A+B)^2-Trace(A+B)\cdot(A+B)+\det (A+B)\cdot I_2=O_2$

but otherwise spot on! Thanks.